The question reads $kx^2 + (k+2)x - 3 = 0$ has roots which are real and positive. Find the possible values k might have.
Now, since it has real and positive roots, the discriminant $\Delta{d} > 0$, so:
$(k+2)^2 - 4(k)(-3) > 0$
=> $k^2 + 4 + 4k + 12k > 0$
=> $k^2 + 16k + 4 > 0$
Now, solving it with the formula, we get:
$\frac{-16 \pm {\sqrt{250}}}{2}$
But when we factorise it, we get:
=> $k(k+16) > -4$
=> $k > -4$ or $k > -20$
BUT, that's NOT the answer, to get the answer, we have to:
$k^2 + 16k + 4 + 64> 64$
=> $k^2 + 16k + 64> 60$
=> $(k+8)^2 > 60$
=> $k > -8 + \sqrt{60}$
That is the answer.
My question is how is this possible that I am getting three different answers and why is the last one correct?
$f(0)<0$ if $k>0$, f(x) has positive root. if $ k<0$, need $ -\frac{k+2}{k}>0$ →$k>-2$. So the answer is $$k>-2, (k≠0)$$